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properties of the Lebesgue integral of nonnegative measurable functions

Theorem.

Let (X,?,μ)X?μ(X,\mathfrak{B},\mu) be a measure space, f:X→[0,∞]normal-:fnormal-→X0f\colon X\to[0,\infty] and g:X→[0,∞]normal-:gnormal-→X0g\colon X\to[0,\infty] be measurable functions, and A,B∈?AB?A,B\in\mathfrak{B}. Then the following properties hold:

1.

∫Af⁢d⁢μ≥0subscriptAfdμ0\displaystyle\int_{A}f\,d\mu\geq 0

2.

If f≤gfgf\leq g, then ∫Af⁢d⁢μ≤∫Ag⁢d⁢μsubscriptAfdμsubscriptAgdμ\displaystyle\int_{A}f\,d\mu\leq\int_{A}g\,d\mu.

3.

∫Af⁢d⁢μ=∫XχA⁢f⁢d⁢μsubscriptAfdμsubscriptXsubscriptχAfdμ\displaystyle\int_{A}f\,d\mu=\int_{X}\chi_{A}f\,d\mu, where χAsubscriptχA\chi_{A} denotes the characteristic function of AAA

4.

If A⊆BABA\subseteq B, then ∫Af⁢d⁢μ≤∫Bf⁢d⁢μsubscriptAfdμsubscriptBfdμ\displaystyle\int_{A}f\,d\mu\leq\int_{B}f\,d\mu.

5.

If c≥0c0c\geq 0, then ∫Ac⁢f⁢d⁢μ=c⁢∫Af⁢d⁢μsubscriptAcfdμcsubscriptAfdμ\displaystyle\int_{A}cf\,d\mu=c\int_{A}f\,d\mu.

6.

If μ⁢(A)=0μA0\mu(A)=0, then ∫Af⁢d⁢μ=0subscriptAfdμ0\displaystyle\int_{A}f\,d\mu=0.

If A∩B=∅ABA\cap B=\emptyset, then ∫A∪Bf⁢d⁢μ=∫Af⁢d⁢μ+∫Bf⁢d⁢μsubscriptABfdμsubscriptAfdμsubscriptBfdμ\displaystyle\int_{{A\cup B}}f\,d\mu=\int_{A}f\,d\mu+\int_{B}f\,d\mu.

9.

If f=gfgf=g almost everywhere with respect to μμ\mu, then ∫Af⁢d⁢μ=∫Ag⁢d⁢μsubscriptAfdμsubscriptAgdμ\displaystyle\int_{A}f\,d\mu=\int_{A}g\,d\mu.

Proof.

1.

Let sss be a simple function with 0≤s≤f0sf0\leq s\leq f. Let s=∑k=1nck⁢χAkssuperscriptsubscriptk1nsubscriptcksubscriptχsubscriptAk\displaystyle s=\sum_{{k=1}}^{n}c_{k}\chi_{{A_{k}}} for ck∈[0,∞]subscriptck0c_{k}\in[0,\infty] and Ak∈?subscriptAk?A_{k}\in\mathfrak{B}. Then ∫As⁢d⁢μ=∑k=1nck⁢μ⁢(A∩Ak)≥0subscriptAsdμsuperscriptsubscriptk1nsubscriptckμAsubscriptAk0\displaystyle\int_{A}s\,d\mu=\sum_{{k=1}}^{n}c_{k}\mu(A\cap A_{k})\geq 0. By definition, ∫Af⁢d⁢μ≥∫As⁢d⁢μsubscriptAfdμsubscriptAsdμ\displaystyle\int_{A}f\,d\mu\geq\int_{A}s\,d\mu. It follows that ∫Af⁢d⁢μ≥0subscriptAfdμ0\displaystyle\int_{A}f\,d\mu\geq 0.

2.

Let sss be a simple function with 0≤s≤f0sf0\leq s\leq f. Since f≤gfgf\leq g, 0≤s≤g0sg0\leq s\leq g. By definition, ∫As⁢d⁢μ≤∫Ag⁢d⁢μsubscriptAsdμsubscriptAgdμ\displaystyle\int_{A}s\,d\mu\leq\int_{A}g\,d\mu. Since this holds for every simple function sss with 0≤s≤f0sf0\leq s\leq f, it follows that ∫Af⁢d⁢μ≤∫Ag⁢d⁢μsubscriptAfdμsubscriptAgdμ\displaystyle\int_{A}f\,d\mu\leq\int_{A}g\,d\mu.

3.

Let sss be a simple function with 0≤s≤f0sf0\leq s\leq f. Then 0≤χA⁢s≤χA⁢f0subscriptχAssubscriptχAf0\leq\chi_{A}s\leq\chi_{A}f. Let s=∑k=1nck⁢χAkssuperscriptsubscriptk1nsubscriptcksubscriptχsubscriptAk\displaystyle s=\sum_{{k=1}}^{n}c_{k}\chi_{{A_{k}}} for ck∈[0,∞]subscriptck0c_{k}\in[0,\infty] and Ak∈?subscriptAk?A_{k}\in\mathfrak{B}. Then